How Fast Must the Cylinder Rotate to Keep Passengers from Sliding?

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The discussion revolves around calculating the minimum rotational speed required for passengers in a rotating cylinder to avoid sliding when the floor drops. Key equations involve the relationship between the coefficient of friction, gravitational force, and centripetal acceleration. The static coefficient of friction for clothing against steel ranges from 0.60 to 1.0, influencing the calculations. A participant initially calculated a frequency of 0.4068 rotations per second, but it was pointed out that this value needed conversion to rotations per minute for accuracy. The conversation emphasizes the importance of correctly applying the coefficient of friction and ensuring unit consistency in calculations.
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Homework Statement



In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will “stick” to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.60 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70.

Homework Equations



T = 1/f

and so the people don't slide

(coefficient of friction) (Fn)= mg

( (coefficient of friction)*m*v^2)/ r = mg



The Attempt at a Solution



(coefficient of friction) = (g*r)/ (v^2)

T = (2*pi*r)/ square root of ((g*r)/(coefficient of friction))

and f= 1/T

I got 0.4068 = f
but the program says I am wrong
 
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What did you use for the coefficient of friction? What did you use for the radius?
 
Assuming that the question was to find the minimum frequency to insure that they wouldn't slide, it looks OK to me. How was the question phrased exactly?
 
The 0.4068 is rotations in a second. You had to multiply that value by 60 to put it into rotations per MINUTE.
 
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